Page 293 of Rotman's Introduction to Homological Algebra:
$f^{*}: \mathbf{pSh}(X, \mathbf{Ab}) \to \mathbf{pSh}(Y, \mathbf{Ab})$ called inverse image,
with $f^{*}: P \mapsto \Gamma(\cdot,\ {P^e}^t)$.
I don't quite get this definition here as it seems different from the normal one using colimit. Any help would be appreciated!
If you mean the definition via etale spaces, I think it should be $f^\ast: P\mapsto \Gamma( -, Y\times_X P^\mathrm{et})$.
I guess the comparison with another definition you have in mind is $$\Gamma(U, Y\times_X P^\mathrm{et})\cong \mathrm{colim}_{f(U)\subset V} P(V).$$ You write $f^\ast$ as a functor between presheaf categories, but this is true only when $P$ is a sheaf, because when $f=\mathrm{id}_X$, the RHS is $P(U)$ while the LHS is $\tilde{P}(U)$, where $\tilde{P}$ is the sheafification of $P$. When $P$ is a sheaf, one way to prove the canonical isomorphism is to say that they enjoy the same universal property, but you can also argue directly as follows:
By the definition of pullbacks, the left-hand side is canonically isomorphic to the abelian group of continuous sections $f(U)\to P^\mathrm{et}$ (which I abuse the notation and denote by $\Gamma(f(U), P^\mathrm{et})$), so it suffices to prove that, for a (not necessarily open) subset $A\subset Y$, we have a canonical isomorphism $\Gamma(A, P^\mathrm{et})\cong \mathrm{colim}_{A\subset V} P(V)$. Now, for any $A\subset V$, the restriction defines $P(V)\to \Gamma(A, P^\mathrm{et})$. Since these form a compatible family of maps, it defines a map $$\mathrm{colim}_{A\subset V} P(V)\to \Gamma(A, P^\mathrm{et}).$$ To prove the bijectivity when $P$ is a sheaf, you need to go back to the definition of the etale space: the RHS is a subset of $\prod_{x\in A}P_x$ consisting of continuous sections. To prove the injectivity, take $s_1\in P(V_1)$ and $s_2\in P(V_2)$, such that $(s_1)_x = (s_2)_x$ for all $x\in A$. This means that there exists an open neighborhood $U_x\subset V_1\cap V_2$ of $x$ such that $s_1$ and $s_2$ restricts to the same sections on $U_x$. Let $U = \bigcup_{x\in A} U_x$. Then $A\subset U\subset V_1\cap V_2$ and $s_1|_U = s_2|_U$, so $s_1$ and $s_2$ represent the same element in the colimit. The proof of surjectivity is similar: take a continuous section $s: A\to P^\mathrm{et}$. By definition, we have $s(x)\in P_x$, and the continuity means that there exists a neighborhood $U_x$ of $x$ and a section $t(x)\in P(U_x)$ such that $s(y) = t(x)_y$ for all $y\in U_x$. Now because $t(x_1)$ and $t(x_2)$ agree on stalks (and thus on the intersection $U_{x_1}\cap U_{x_2}$), they paste together to define a section $t\in P(U)$, where $U = \bigcup_{x\in A}U_x\supset A$. Now the surjectivity follows because $s\in\Gamma(A, P^\mathrm{et})$ is the image of $t\in P(U)$.